The Quotient Rule - Problem 1
Let's do some problems with the quotient rule. So let's just recall that the quotient rule is how we differentiate a quotient of two functions; f and g. We'll call f the high function, and g the low function. Remember that the quotient rule is low d high minus high d low over the square of what's below.
So let's take a look at this problem. We want to differentiate y equals e to the x over x². So dy/dx, the derivative, is going to be low d high. So I need to put my fraction bar in. Low d high would be x² times the derivative of e to the x which is e to the x, minus high d low that's e to the x times the derivative of this guy which is 2x, over the square of what's below. That becomes x to the fourth.
This can be simplified a little bit. I can pull an e to the x out. I'm left with x² minus 2x. Now one of those x's will cancel with the denominator. So this x, this x, and that guy. So I'm left with an x minus 2 all over x³. So there is your answer.
Let's try another one. I have y equals x² plus 3x plus 2 over 5 minus x. Again, the derivative is low d high. So this is the low function. D high would be 2x plus 3 minus high d low. The high function is x² plus 3x plus 2. D low would be -1, the derivative of 5 minus x, -1 over the square of what's below. So 5 minus x².
I can simplify this. I don't want to, it looks terrible. 5 minus x, let's see we're going to get -2x². We're going to get 10x minus 3x. So plus 7x and then we're going to get +50. Then here, we'll get minus, minus plus x². We're going to get plus 3x. We're going to get plus 2 all over 5 minus x². I'll leave the denominator that way.
Just to combine like terms in the numerator; I'm going to have minus x². I'm going to have 10x, and I'm going to have plus 17 all that over (5 minus x)². That's the quotient rule.